Question

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Answer 1

LHS :

tan4x=tan[2(2x)]

=1−tan22x2tan2x

=1−(1−tan2x2tanx)22(1−tan2x2tanx)

=(1−tan2x)2−4tan2x4tanx(1−tan2x)

=1−6tan2x+tan4x4tanx(1−tan2x) 

= RHS

Answer 2

Answer:  LHS :

tan4x=tan[2(2x)]

=1−tan22x2tan2x

=1−(1−tan2x2tanx)22(1−tan2x2tanx)

=(1−tan2x)2−4tan2x4tanx(1−tan2x)

=1−6tan2x+tan4x4tanx(1−tan2x)  

= RHS

Step-by-step explanation: